Frame AFD, SFD, BMD (40 kN at 45 deg)

Assumptions (taken from the figure):

• A(0,0) hinged at the base; AB = 4 m, with an internal hinge at its top B; horizontal triangular load from 20 kN/m at A to 0 at B (→), total 40 kN acting 1.33 m above A.
• Beam B–C–D–E: BC = 1 m, CD = 1 m, DE = 2 m; 30 kN downward at C; 10 kN/m downward over DE (20 kN); 40 kN at 45° (down and to the right) at E, with components 28.28 kN → and 28.28 kN ↓.
• The inclined member EF descends 4 m over 1 m horizontally to the hinged support F = (5, 0).

Sign convention: N is positive in tension; V is positive when the forces on the part before the section (travelling A → end) point along the left-hand normal of the member axis (upward on a left-to-right beam); M is quoted as a magnitude with the tension face named. Support reactions are drawn on the free-body of the whole frame.

Bo----C------D------------E
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A                               F

Support reactions

• Hinge at B (moments of column AB about B): $4H_A + 40(4 - 1.33) = 0 \Rightarrow H_A = -26.67$ kN.
• Moments about A: $5V_F = 40(1.33) + 30(1) + 20(3) + 28.28(4) + 28.28(4) = 369.61 \Rightarrow V_F = 73.92$ kN.
• $\Sigma F_y = 0$: $V_A = 30 + 20 + 28.28 - 73.92 = 4.36$ kN.
• $\Sigma F_x = 0$: $H_F = -(40 + 28.28) + 26.67 = -41.62$ kN.

Results:

• $H_A$ = 26.67 kN ←
• $V_A$ = 4.36 kN ↑
• $H_F$ = 41.62 kN ←
• $V_F$ = 73.92 kN ↑

Check: moments of EF about E from F: $1(73.92) - 4(41.62) = -92.55$ kN·m, which matches the beam moment at E below ✓

Axial force, shear force and bending moment at the key sections

Shape of the diagrams

• AB: V parabolic, M cubic
• BC: V constant, M linear
• CD: V constant, M linear
• DE: V linear, M parabolic
• EF: V constant, M linear
• AFD: constant between loads on a member, changing only where a load component acts along the member.
• The BMD is drawn on the tension side of each member; it is continuous at rigid joints (moment is transferred) and zero at hinges and at free ends.

Salient features

• Maximum bending moment = 92.55 kN·m at joint E (top of DE, right face of EF)
• V = 0 at 1.69 m from A along AB: M = 20.53 kN·m (right face), the local maximum
• Maximum sagging moment on the beam = 4.36 kN·m at C
• Point of contraflexure (M = 0) at 1.17 m from B along the beam (0.17 m to the right of C)
• The bending moment is zero at the internal hinge B (from both sides).